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A change in water temperature with no change of state requires one calorie per gram of pure water per degree Celsius.

90Â° - 15Â° = 75Â°

80 g x 75Â° C x 1 cal/g/Â°C = 6000 cal = 6 kcal

Q: How much heat is needed to raise temp of 80g of water from 15 degrees celsius to 90 degrees celsius?

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Heating of water=m x s x delta T,where m is the mass ,s is the specific heat of water(1 cal/gm)=5x1x(50-25) =125 cal

8.200 J

The energy required to raise the temperature 1 degree Celsius of 1 gram of water (1 mL) is 1 calorie (=4.18 J). So for 1 kg, 1Kcal (= 4180 J = 4.18 KJ) is required. To raise it 60 degrees, just multiply by 60 and for 10 kg multiply by 10 again. That would make 2.508 MJ (= 2508000 J) Now this is not completely accurate. The energy required to raise the temperature of water differs at 20 degrees from that at 60 degrees. The difference is small (~0.05 J or something like that) but still present.

This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).

32 Degrees Fahrenheit 0 Degrees Celsius

Related questions

2,641,760J...

2,641,760J...

The number of calories required will depend on the mass of water which is to be heated.

42 J

there are no calories in water you idiot

(5)(3)= 15 calories. 1 calorie is the energy (heat) to raise 1 gram of water by 1 degree celsius, so 5 grams of water (3 degrees Celsius) = 15.

it can be made to boil at 105 degrees Celsius if we add impurities to it,as impurities raise the boiling point.

Q=mcÎ”T Q=14 x 4200 x 21.6 Q=1270080J

If its in Celsius then another 13 degrees are needed because water boils at 100 degrees Celsius

4,200 J/kgC (83-4) x 8kg x 4,200 = 2,654,400 joules

12 degrees Celsius

12 degrees Celsius